Chapter – 2: Relations and Function
Exercise 2.1
1. If
, find the values of x and y.
Solution: –
Given,
As the ordered pairs are equal, the corresponding elements should also be equal.
Thus,
x/3 + 1 = 5/3 and y – 2/3 = 1/3
Solving, we get
x + 3 = 5 and 3y – 2 = 1 [Taking L.C.M and adding]
x = 2 and 3y = 3
Therefore,
x = 2 and y = 1
2. If the set A has 3 elements and the set B = {3, 4, 5}, then find the number of elements in (A × B)?
Solution: –
Given, set A has 3 elements and the elements of set B are {3, 4, and 5}.
So, the number of elements in set B = 3
Then, the number of elements in (A × B) = (Number of elements in A) × (Number of elements in B)
= 3 × 3 = 9
Therefore, the number of elements in (A × B) will be 9.
3. If G = {7, 8} and H = {5, 4, 2}, find G × H and H × G.
Solution: –
Given, G = {7, 8} and H = {5, 4, 2}
We know that,
The Cartesian product of two non-empty sets P and Q is given as
P × Q = {(p, q): p ∈ P, q ∈ Q}
So,
G × H = {(7, 5), (7, 4), (7, 2), (8, 5), (8, 4), (8, 2)}
H × G = {(5, 7), (5, 8), (4, 7), (4, 8), (2, 7), (2, 8)}
4. State whether each of the following statements are true or false. If the statement is false, rewrite the given statement correctly.
(i) If P = {m, n} and Q = {n, m}, then P × Q = {(m, n), (n, m)}.
(ii) If A and B are non-empty sets, then A × B is a non-empty set of ordered pairs (x, y) such that x ∈ A and y ∈ B.
(iii) If A = {1, 2}, B = {3, 4}, then A × (B ∩ Φ) = Φ.
Solution: –
(i) The statement is False. The correct statement is:
If P = {m, n} and Q = {n, m}, then
P × Q = {(m, m), (m, n), (n, m), (n, n)}
(ii) True
(iii) True
5. If A = {–1, 1}, find A × A × A.
Solution: –
The A × A × A for a non-empty set A is given by
A × A × A = {(a, b, c): a, b, c ∈ A}
Here, It is given A = {–1, 1}
So,
A × A × A = {(–1, –1, –1), (–1, –1, 1), (–1, 1, –1), (–1, 1, 1), (1, –1, –1), (1, –1, 1), (1, 1, –1), (1, 1, 1)}
6. If A × B = {(a, x), (a, y), (b, x), (b, y)}. Find A and B.
Solution: –
Given,
A × B = {(a, x), (a, y), (b, x), (b, y)}
We know that the Cartesian product of two non-empty sets P and Q is given by:
P × Q = {(p, q): p ∈ P, q ∈ Q}
Hence, A is the set of all first elements and B is the set of all second elements.
Therefore, A = {a, b} and B = {x, y}
7. Let A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}. Verify that
(i) A × (B ∩ C) = (A × B) ∩ (A × C)
(ii) A × C is a subset of B × D
Solution: –
Given,
A = {1, 2}, B = {1, 2, 3, 4}, C = {5, 6} and D = {5, 6, 7, 8}
(i) To verify: A × (B ∩ C) = (A × B) ∩ (A × C)
Now, B ∩ C = {1, 2, 3, 4} ∩ {5, 6} = Φ
Thus,
L.H.S. = A × (B ∩ C) = A × Φ = Φ
Next,
A × B = {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (2, 4)}
A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}
Thus,
R.H.S. = (A × B) ∩ (A × C) = Φ
Therefore, L.H.S. = R.H.S
– Hence verified
(ii) To verify: A × C is a subset of B × D
First,
A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}
And,
B × D = {(1, 5), (1, 6), (1, 7), (1, 8), (2, 5), (2, 6), (2, 7), (2, 8), (3, 5), (3, 6), (3, 7), (3, 8), (4, 5), (4, 6), (4, 7), (4, 8)}
Now, it’s clearly seen that all the elements of set A × C are the elements of set B × D.
Thus, A × C is a subset of B × D.
– Hence verified
8. Let A = {1, 2} and B = {3, 4}. Write A × B. How many subsets will A × B have? List them.
Solution: –
Given,
A = {1, 2} and B = {3, 4}
So,
A × B = {(1, 3), (1, 4), (2, 3), (2, 4)}
Number of elements in A × B is n(A × B) = 4
We know that,
If C is a set with n(C) = m, then n[P(C)] = 2m.
Thus, the set A × B has 24 = 16 subsets.
And, these subsets are as below:
Φ, {(1, 3)}, {(1, 4)}, {(2, 3)}, {(2, 4)}, {(1, 3), (1, 4)}, {(1, 3), (2, 3)}, {(1, 3), (2, 4)}, {(1, 4), (2, 3)}, {(1, 4), (2, 4)}, {(2, 3), (2, 4)}, {(1, 3), (1, 4), (2, 3)}, {(1, 3), (1, 4), (2, 4)}, {(1, 3), (2, 3), (2, 4)}, {(1, 4), (2, 3), (2, 4)}, {(1, 3), (1, 4), (2, 3), (2, 4)}
9. Let A and B be two sets such that n(A) = 3 and n (B) = 2. If (x, 1), (y, 2), (z, 1) are in A × B, find A and B, where x, y and z are distinct elements.
Solution: –
Given,
n(A) = 3 and n(B) = 2; and (x, 1), (y, 2), (z, 1) are in A × B.
We know that,
A = Set of first elements of the ordered pair elements of A × B
B = Set of second elements of the ordered pair elements of A × B.
So, clearly x, y, and z are the elements of A; and
1 and 2 are the elements of B.
As n(A) = 3 and n(B) = 2, it is clear that set A = {x, y, z} and set B = {1, 2}.
10. The Cartesian product A × A has 9 elements among which are found (–1, 0) and (0, 1). Find the set A and the remaining elements of A × A.
Solution: –
We know that,
If n(A) = p and n(B) = q, then n(A × B) = pq.
Also, n(A × A) = n(A) × n(A)
Given,
n(A × A) = 9
So, n(A) × n(A) = 9
Thus, n(A) = 3
Also given that, the ordered pairs (–1, 0) and (0, 1) are two of the nine elements of A × A.
And, we know in A × A = {(a, a): a ∈ A}.
Thus, –1, 0, and 1 has to be the elements of A.
As n(A) = 3, clearly A = {–1, 0, 1}.
Hence, the remaining elements of set A × A are as follows:
(–1, –1), (–1, 1), (0, –1), (0, 0), (1, –1), (1, 0), and (1, 1)
Exercise 2.2
1. Let A = {1, 2, 3, … , 14}. Define a relation R from A to A by R = {(x, y): 3x – y = 0, where x, y ∈ A}. Write down its domain, codomain and range.
Solution: –
The relation R from A to A is given as:
R = {(x, y): 3x – y = 0, where x, y ∈ A}
= {(x, y): 3x = y, where x, y ∈ A}
So,
R = {(1, 3), (2, 6), (3, 9), (4, 12)}
Now,
The domain of R is the set of all first elements of the ordered pairs in the relation.
Hence, Domain of R = {1, 2, 3, 4}
The whole set A is the codomain of the relation R.
Hence, Codomain of R = A = {1, 2, 3, …, 14}
The range of R is the set of all second elements of the ordered pairs in the relation.
Hence, Range of R = {3, 6, 9, 12}
2. Define a relation R on the set N of natural numbers by R = {(x, y): y = x + 5, x is a natural number less than 4; x, y ∈ N}. Depict this relationship using roster form. Write down the domain and the range.
Solution: –
The relation R is given by:
R = {(x, y): y = x + 5, x is a natural number less than 4, x, y ∈ N}
The natural numbers less than 4 are 1, 2, and 3.
So,
R = {(1, 6), (2, 7), (3, 8)}
Now,
The domain of R is the set of all first elements of the ordered pairs in the relation.
Hence, Domain of R = {1, 2, 3}
The range of R is the set of all second elements of the ordered pairs in the relation.
Hence, Range of R = {6, 7, 8}
3. A = {1, 2, 3, 5} and B = {4, 6, 9}. Define a relation R from A to B by R = {(x, y): the difference between x and y is odd; x ∈ A, y ∈ B}. Write R in roster form.
Solution: –
Given,
A = {1, 2, 3, 5} and B = {4, 6, 9}
The relation from A to B is given as:
R = {(x, y): the difference between x and y is odd; x ∈ A, y ∈ B}
Thus,
R = {(1, 4), (1, 6), (2, 9), (3, 4), (3, 6), (5, 4), (5, 6)}
4. The figure shows a relationship between the sets P and Q. write this relation
(i) in set-builder form (ii) in roster form.
What is its domain and range?
Solution: –
From the given figure, it’s seen that
P = {5, 6, 7}, Q = {3, 4, 5}
The relation between P and Q:
Set-builder form
(i) R = {(x, y): y = x – 2; x ∈ P} or R = {(x, y): y = x – 2 for x = 5, 6, 7}
Roster form
(ii) R = {(5, 3), (6, 4), (7, 5)}
Domain of R = {5, 6, 7}
Range of R = {3, 4, 5}
5. Let A = {1, 2, 3, 4, 6}. Let R be the relation on A defined by
{(a, b): a, b ∈ A, b is exactly divisible by a}.
(i) Write R in roster form
(ii) Find the domain of R
(iii) Find the range of R.
Solution: –
Given,
A = {1, 2, 3, 4, 6} and relation R = {(a, b): a, b ∈ A, b is exactly divisible by a}
Hence,
(i) R = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 6), (2, 2), (2, 4), (2, 6), (3, 3), (3, 6), (4, 4), (6, 6)}
(ii) Domain of R = {1, 2, 3, 4, 6}
(iii) Range of R = {1, 2, 3, 4, 6}
6. Determine the domain and range of the relation R defined by R = {(x, x + 5): x ∈ {0, 1, 2, 3, 4, 5}}.
Solution: –
Given,
Relation R = {(x, x + 5): x ∈ {0, 1, 2, 3, 4, 5}}
Thus,
R = {(0, 5), (1, 6), (2, 7), (3, 8), (4, 9), (5, 10)}
So,
Domain of R = {0, 1, 2, 3, 4, 5} and,
Range of R = {5, 6, 7, 8, 9, 10}
7. Write the relation R = {(x, x3): x is a prime number less than 10} in roster form.
Solution: –
Given,
Relation R = {(x, x3): x is a prime number less than 10}
The prime numbers less than 10 are 2, 3, 5, and 7.
Therefore,
R = {(2, 8), (3, 27), (5, 125), (7, 343)}
8. Let A = {x, y, z} and B = {1, 2}. Find the number of relations from A to B.
Solution: –
Given, A = {x, y, z} and B = {1, 2}.
Now,
A × B = {(x, 1), (x, 2), (y, 1), (y, 2), (z, 1), (z, 2)}
As n(A × B) = 6, the number of subsets of A × B will be 26.
Thus, the number of relations from A to B is 26.
9. Let R be the relation on Z defined by R = {(a, b): a, b ∈ Z, a – b is an integer}. Find the domain and range of R.
Solution: –
Given,
Relation R = {(a, b): a, b ∈ Z, a – b is an integer}
We know that the difference between any two integers is always an integer.
Therefore,
Domain of R = Z and Range of R = Z
Exercise 2.3
1. Which of the following relations are functions? Give reasons. If it is a function, determine its domain and range.
(i) {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}
(ii) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}
(iii) {(1, 3), (1, 5), (2, 5)}
Solution: –
(i) {(2, 1), (5, 1), (8, 1), (11, 1), (14, 1), (17, 1)}
As 2, 5, 8, 11, 14, and 17 are the elements of the domain of the given relation having their unique images, this relation can be called as a function.
Here, domain = {2, 5, 8, 11, 14, 17} and range = {1}
(ii) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5), (12, 6), (14, 7)}
As 2, 4, 6, 8, 10, 12, and 14 are the elements of the domain of the given relation having their unique images, this relation can be called as a function.
Here, domain = {2, 4, 6, 8, 10, 12, 14} and range = {1, 2, 3, 4, 5, 6, 7}
(iii) {(1, 3), (1, 5), (2, 5)}
It’s seen that the same first element i.e., 1 corresponds to two different images i.e., 3 and 5, this relation cannot be called as a function.
2. Find the domain and range of the following real function:
(i) f(x) = –|x| (ii) f(x) = √(9 – x2)
Solution: –
(i) Given,
f(x) = –|x|, x ∈ R
We know that,
As f(x) is defined for x ∈ R, the domain of f is R.
It is also seen that the range of f(x) = –|x| is all real numbers except positive real numbers.
Therefore, the range of f is given by (–∞, 0].
(ii) f(x) = √(9 – x2)
As √(9 – x2) is defined for all real numbers that are greater than or equal to –3 and less than or equal to 3, for 9 – x2 ≥ 0.
So, the domain of f(x) is {x: –3 ≤ x ≤ 3} or [–3, 3].
Now,
For any value of x in the range [–3, 3], the value of f(x) will lie between 0 and 3.
Therefore, the range of f(x) is {x: 0 ≤ x ≤ 3} or [0, 3].
3. A function f is defined by f(x) = 2x – 5. Write down the values of
(i) f(0), (ii) f(7), (iii) f(–3)
Solution: –
Given,
Function, f(x) = 2x – 5.
Therefore,
(i) f(0) = 2 × 0 – 5 = 0 – 5 = –5
(ii) f(7) = 2 × 7 – 5 = 14 – 5 = 9
(iii) f(–3) = 2 × (–3) – 5 = – 6 – 5 = –11
4. The function ‘t’ which maps temperature in degree Celsius into temperature in degree Fahrenheit is defined by.
Find (i) t (0) (ii) t (28) (iii) t (–10) (iv) The value of C, when t(C) = 212
Solution: –
5. Find the range of each of the following functions.
(i) f(x) = 2 – 3x, x ∈ R, x > 0.
(ii) f(x) = x2 + 2, x is a real number.
(iii) f(x) = x, x is a real number.
Solution: –
(i) Given,
f(x) = 2 – 3x, x ∈ R, x > 0.
We have,
x > 0
So,
3x > 0
-3x < 0 [Multiplying by -1 both the sides, the inequality sign changes]
2 – 3x < 2
Therefore, the value of 2 – 3x is less than 2.
Hence, Range = (–∞, 2)
(ii) Given,
f(x) = x2 + 2, x is a real number
We know that,
x2 ≥ 0
So,
x2 + 2 ≥ 2 [Adding 2 both the sides]
Therefore, the value of x2 + 2 is always greater or equal to 2 for x is a real number.
Hence, Range = [2, ∞)
(iii) Given,
f(x) = x, x is a real number
Clearly, the range of f is the set of all real numbers.
Thus,
Range of f = R
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